I have:
$$4x^2y'' + (3x+1)y = 0$$
From this the indicial equation is found so that $$r(r-1)+1/4=0 \Rightarrow r = 1/2 = r_1 = r_2,$$ so the first solution will be of the form:
$$y_1(x) = \sqrt{x}\sum_{k=0}^{\infty}{c_kx^k}$$
($y_2$ follows)
but I need to find the power series solution. I tried substituting it in and solving by finding a recursive relation etc, but I cannot seem to get the answer - am I missing something obvious here, or is it just tedious?
Any tips would be appreciated, I have a test tomorrow and I'm having trouble fully understanding Frobenius even with online resources.
Thanks!
fyi, this is the solution
\begin{eqnarray*} y= \sum_{k=0}^{\infty} c_k x^{k+1/2} \\ \frac{d^2y}{dx^2} = \sum_{k=0}^{\infty}(k+1/2)(k-1/2) c_k x^{k-3/2} \\ \end{eqnarray*} substitute this into the equation gives \begin{eqnarray*} \sum_{k=0}^{\infty}(4k^2-1) c_k x^{k+1/2} +3 \sum_{k=0}^{\infty} c_k x^{k+3/2} +\sum_{k=0}^{\infty} c_k x^{k+1/2} =0\\ \sum_{k=0}^{\infty}(4k^2-1) c_k x^{k+1/2} +3 \sum_{k=1}^{\infty} c_{k-1} x^{k+1/2} +\sum_{k=0}^{\infty} c_k x^{k+1/2} =0\\ \sum_{k=1}^{\infty} (4k^2 c_k +3 c_{k-1}) x^{k+1/2} =0\\ \end{eqnarray*} Thus we have the recurrence $4k^2 c_k +3 c_{k-1}=0$ for $k \geq 1$ & this has solution $c_k =\frac{(-1)^k 3^k}{k!^2 4^k}$ so the first solution is \begin{eqnarray*} y_1(x) = c_0 \sqrt{x}\sum_{k=0}^{\infty} \frac{1}{k!^2 } \left( \frac{-3x}{4} \right)^k. \end{eqnarray*}